Solution
= Solution
Set
$$
u=\log(1+t),
\qquad t=e^u-1,
\qquad dt=e^u\,du.
$$
Then
$$
J(x)=\int_0^{\log(1+\pi)}g(u)e^{-xu}\,du,
\qquad
g(u)=e^u\bigl[1-\cos(e^u-1)\bigr].
$$
The <Taylor series> at the endpoint is
$$
e^u-1=u+\frac{u^2}{2}+\frac{u^3}{6}+O(u^4),
$$
and hence
$$
g(u)=\frac{u^2}{2}+u^3+O(u^4).
$$
The <Watson lemma>, together with
$$
\int_0^\infty u^ne^{-xu}\,du=\frac{n!}{x^{n+1}},
$$
therefore yields
$$
\boxed{J(x)=\frac1{x^3}+\frac6{x^4}+O(x^{-5})}.
$$